Three Dimensional Geometry jee mains questions with solutions

Q 71 :
Let a straight line L pass through the point P(2, –1, 3) and be perpendicular to the lines $\frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{- 2}$ and $\frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4}$ . If the line L intersects the yz-plane at the point Q, then the distance between the points P and Q is : [2025]

$\sqrt{10}$
2
3
$2 \sqrt{3}$

1

2

3

4

(3)

Vector parallel to line $L =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & - 2 \\ 1 & 3 & 4 \end{matrix} |$

$= 10 \hat{i} - 10 \hat{j} + 5 \hat{k} = 5 (2 \hat{i} - 2 \hat{j} + \hat{k})$

Equation of line L passing through point P(2, –1, 3) and parallel to vector $(2 \hat{i} - 2 \hat{j} + \hat{k})$ , is

$\frac{x - 2}{2} = \frac{y + 1}{- 2} = \frac{z - 3}{1} = λ$

$∴ x = 2 λ + 2, y = - 2 λ - 1, z = λ + 3$

$∵$ Line L intersects the yz-plane

$∴ 2 λ + 2 = 0 \Rightarrow λ = - 1$

Hence, point Q is (0, 1, 2)

Distance between point P(2, –1, 3) and Q(0, 1, 2)

$= \sqrt{{(0 - 2)}^{2} + {(1 + 1)}^{2} + {(2 - 3)}^{2}} = \sqrt{9} = 3$ .

Q 72 :
Let the area of the triangle formed by the lines x + 2 = y – 1 = z, $\frac{x - 3}{5} = \frac{y}{- 1} = \frac{z - 1}{1}$ and $\frac{x}{- 3} = \frac{y - 3}{3} = \frac{z - 2}{1}$ be A. Then $A^{2}$ is equal to __________. [2025]

0%

(56)

$L_{1} : x + 2 = y - 1 = z = λ (say)$

$L_{2} : \frac{x - 3}{5} = \frac{y}{- 1} = \frac{z - 1}{1} = μ (say)$

$L_{3} : \frac{x}{- 3} = \frac{y - 3}{3} = \frac{z - 2}{1} = δ (say)$

Any point of line $L_{1}, L_{2} and L_{3}$ is given by $(λ - 2, λ + 1, λ)$ , $(5 μ + 3, - μ, μ + 1)$ and $(- 3 δ, 3 δ + 3, δ + 2)$ respectively.

Point of intersection of $L_{1}$ and $L_{2}$ is given by

$\begin{matrix} λ - 2 = 5 μ + 3 \\ λ + 1 = - μ \\ λ = μ + 1 \end{matrix}} \Rightarrow λ = 0, μ = - 1$

$∴$ Point of intersection is A(–2, 1, 0).

Point of intersection of $L_{2}$ and $L_{3}$ is given by

$\begin{matrix} 5 μ + 3 = - 3 δ \\ - μ = 3 δ + 3 \\ μ + 1 = δ + 2 \end{matrix}} \Rightarrow μ = 0, δ = - 1$

$∴$ Point of intersection is B(3, 0 , 1).

Point of intersection of $L_{1}$ and $L_{3}$ is given by

$\begin{matrix} λ - 2 = - 3 δ \\ λ + 1 = 3 δ + 3 \\ λ = δ + 2 \end{matrix}} \Rightarrow λ = 2, δ = 0$

$∴$ Point of intersection is C(0, 3, 2).

Now, Area of $△ A B C = \frac{1}{2}$ $| \vec{A B} \times \vec{B C} |$

$∴ A = \frac{1}{2}$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 5 & 1 & - 1 \\ - 3 & 3 & 1 \end{matrix} | |$

$= \frac{1}{2} | 4 \hat{i} + 8 \hat{j} - 12 \hat{k |}$

$= \frac{1}{2} \sqrt{16 + 64 + 144} = \sqrt{56}$

$∴ A^{2} = 56$ .

Q 73 :
Let $L_{1} : \frac{x - 1}{3} = \frac{y - 1}{- 1} = \frac{z + 1}{0}$ and $L_{2} : \frac{x - 2}{2} = \frac{y}{0} = \frac{z + 4}{α}, α \in R$ be two lines, which intersect at the point B. If P is the foot of perpendicular from the point A(1, 1, –1) and $L_{2}$ , then the value of $26 α {(P B)}^{2}$ is __________. [2025]

0%

(216)

Point $B \equiv (3 μ + 1, - μ + 1, - 1) \equiv (2 λ + 2, 0, α λ - 4)$

$\Rightarrow 3 μ + 1 = 2 λ + 2, - μ + λ = 0, λ α - 4 = - 1$

$∴ μ = 1 \Rightarrow λ = 1 and α = 3$

So, point B(4, 0, –1).

Let point P is (2k + 2, 0, 3k – 4).

So, Dr's of AP is < 2k + 1, –1, 3k – 3 >

Since, $L_{2} ⊥ A P$

$\Rightarrow 2 (2 k + 1) + 0 (- 1) + 3 (3 k - 3) = 0$

$\Rightarrow 4 k + 2 + 9 k - 9 = 0 \Rightarrow k = \frac{7}{13}$

$∴ Point P \equiv (\frac{40}{13}, 0, \frac{- 31}{13})$

$∴ {(P B)}^{2} = {(\frac{40}{13} - 4)}^{2} + {(0 - 0)}^{2} + {(\frac{- 31}{13} + 1)}^{2} = \frac{468}{169}$

$∴ 26 α {(P B)}^{2} = 26 \times 3 \times \frac{468}{169} = 216$ .

Q 74 :
Let P be the image of the point Q(7, –2, 5) in the line $L : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and R(5, p, q) be a point on L. Then the square of the area of $∆ PQR$ is __________. [2025]

0%

(957)

Given, $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and R(5, p, q) be on the line.

Here P be the image of point.

Since, R is on the line L, then

$R \equiv (2 λ + 1, 3 λ - 1, 4 λ) = (5, p, q) [Given]$

$\Rightarrow 2 λ + 1 = 5$

$\Rightarrow 2 λ = 4 \Rightarrow λ = 2$

$∴ R \equiv (5, 5, 8)$

Since, T is also on the line L, then

$T \equiv (2 μ + 1, 3 μ - 1, 4 μ)$

Now, $\vec{Q T} = (2 μ - 6) \hat{i} + (3 μ + 1) \hat{j} + (4 μ - 5) \hat{k}$

and $\vec{b} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ (Normal)

Taking $\vec{Q T} \cdot \vec{b} = 0$

$\Rightarrow 4 μ - 12 + 9 μ + 3 + 16 μ - 20 = 0 \Rightarrow 29 μ - 29 = 0$

$\Rightarrow μ = 1$

$∴ T \equiv (3, 2, 4)$

$\Rightarrow Q T = \sqrt{{(7 - 3)}^{2} + {(- 2 - 2)}^{2} + {(5 - 4)}^{2}}$

$= \sqrt{16 + 16 + 1} = \sqrt{33} \Rightarrow P Q = 2 \sqrt{33}$

Similarly, $R T = \sqrt{29}$

$\Rightarrow Area of △ P Q R = \frac{1}{2} \times \sqrt{29} \times 2 \sqrt{33} = \sqrt{29} \times \sqrt{33}$

$\Rightarrow {(Area of △ P Q R)}^{2} = {(\sqrt{29} \times \sqrt{33})}^{2} = 29 \times 33 = 957$ .

Topic Question Set

Q 72 :
Let the area of the triangle formed by the lines x + 2 = y – 1 = z, $\frac{x - 3}{5} = \frac{y}{- 1} = \frac{z - 1}{1}$ and $\frac{x}{- 3} = \frac{y - 3}{3} = \frac{z - 2}{1}$ be A. Then $A^{2}$ is equal to __________. [2025]

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Q 74 :
Let P be the image of the point Q(7, –2, 5) in the line $L : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and R(5, p, q) be a point on L. Then the square of the area of $∆ PQR$ is __________. [2025]

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Topic Question Set

Q 71 : Let a straight line L pass through the point P(2, –1, 3) and be perpendicular to the lines x–12=y+11=z–3–2 and x–31=y–23=z+24. If the line L intersects the yz-plane at the point Q, then the distance between the points P and Q is : [2025]

Q 72 : Let the area of the triangle formed by the lines x + 2 = y – 1 = z, x–35=y–1=z–11 and x–3=y–33=z–21 be A. Then A2 is equal to __________. [2025]

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Q 73 : Let L1:x–13=y–1–1=z+10 and L2:x–22=y0=z+4α, α∈R be two lines, which intersect at the point B. If P is the foot of perpendicular from the point A(1, 1, –1) and L2, then the value of 26α(PB)2 is __________. [2025]

0%

Q 74 : Let P be the image of the point Q(7, –2, 5) in the line L:x–12=y+13=z4 and R(5, p, q) be a point on L. Then the square of the area of ∆PQR is __________. [2025]

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Q 72 :
Let the area of the triangle formed by the lines x + 2 = y – 1 = z, $\frac{x - 3}{5} = \frac{y}{- 1} = \frac{z - 1}{1}$ and $\frac{x}{- 3} = \frac{y - 3}{3} = \frac{z - 2}{1}$ be A. Then $A^{2}$ is equal to __________. [2025]

Q 74 :
Let P be the image of the point Q(7, –2, 5) in the line $L : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and R(5, p, q) be a point on L. Then the square of the area of $∆ PQR$ is __________. [2025]